On a sunny day, a tower casts a shadow 35.2m long. At the same time, a 1.3m parking meter that is nearby casts a shadow 1.8m long. How high is the tower to the nearest tenth of a metre?

Triangle formed by parking meter:

$\displaystyle c^2 = a^2 + b^2$

$\displaystyle c^2 = 1.3^2 + 1.8^2$

$\displaystyle c^2 = \frac{493}{100}$

$\displaystyle c = \frac{\sqrt{493}}{10}$

$\displaystyle SinA = \frac{O}{H}$

$\displaystyle SinA = \frac{1.3}{\frac{\sqrt{493}}{10}}$

$\displaystyle Sin^{-1}\frac{1.3}{\frac{\sqrt{493}}{10}} = A$

$\displaystyle 35.84^{\circ} = A$

Triangle formed by tower (Total triangle):

$\displaystyle \frac{a}{SinA} = \frac{b}{SinB}$

$\displaystyle \frac{a}{Sin35.84} = \frac{37}{Sin54.16}$

$\displaystyle a = Sin35.84(\frac{37}{Sin54.16})$

$\displaystyle a = 26.7m$

*The $\displaystyle 37m$ value was obtained by adding $\displaystyle 35.2m$ and $\displaystyle 1.8m$.

*The $\displaystyle 54.16^{\circ}$ value was obtained by subtracting $\displaystyle 90deg$ and $\displaystyle 35.84deg$ from $\displaystyle 180deg$.

Therefore, the height of the tower is $\displaystyle 26.7m$.

Textbook answer: $\displaystyle 25.4m$.

Where did I make the mistake?